Write the blue letters that are inside closed figures. These letters can be arranged into two words that tell the ride that Rach
1 answer:
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Answer:
(a) Null Hypothesis, :
= 22 ounces
Alternate Hypothesis, :
22 ounces
(b) The value of the test statistic is -2.687.
(c) The critical values are -1.96 and 1.96.
(d) We conclude that Reject since the value of the test statistic is less than the negative critical value.
Step-by-step explanation:
We are given that a local bottler in Hawaii wishes to ensure that an average of 22 ounces of passion fruit juice is used to fill each bottle.
He takes a random sample of 65 bottles. The mean weight of the passion fruit juice in the sample is 21.54 ounces. Assume that the population standard deviation is 1.38 ounce.
<em>Let </em><em> = average ounces of passion fruit juice used to fill each bottle.</em>
(a) So, Null Hypothesis, :
= 22 ounces {means that the average of 22 ounces of passion fruit juice is used to fill each bottle}
Alternate Hypothesis, :
22 ounces {means that the average different from 22 ounces of passion fruit juice is used to fill each bottle}
The test statistics that would be used here <u>One-sample z test statistics</u> as we know about the population standard deviation;
T.S. = ~ N(0,1)
where, = sample mean weight of the passion fruit juice = 21.54 ounces
= population standard deviation = 1.38 ounce
n = sample of bottles = 65
So, <u><em>test statistics</em></u> =
= -2.687
(b) The value of z test statistics is -2.687.
(c) Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.
<em>Since our test statistics does not lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><u><em>we reject our null hypothesis</em></u><em>.</em>
Therefore, we conclude that Reject since the value of the test statistic is less than the negative critical value which means that the average different from 22 ounces of passion fruit juice is used to fill each bottle.
The valid conclusions for the manager based on the considered test is given by: Option
<h3>When do we perform one sample z-test?</h3>One sample z-test is performed if the sample size is large enough (n > 30) and we want to know if the sample comes from the specific population.
For this case, we're specified that:
- Population mean =
= $150
- Population standard deviation =
= $30.20
- Sample mean =
= $160
- Sample size = n = 40 > 30
- Level of significance =
= 2.5% = 0.025
- We want to determine if the average customer spends more in his store than the national average.
Forming hypotheses:
- Null Hypothesis: Nullifies what we're trying to determine. Assumes that the average customer doesn't spend more in the store than the national average. Symbolically, we get:
- Alternate hypothesis: Assumes that customer spends more in his store than the national average. Symbolically
where is the hypothesized population mean of the money his customer spends in his store.
The z-test statistic we get is:
The test is single tailed, (right tailed).
The critical value of z at level of significance 0.025 is 1.96
Since we've got 2.904 > 1.96, so we reject the null hypothesis.
(as for right tailed test, we reject null hypothesis if the test statistic is > critical value).
Thus, we accept the alternate hypothesis that customer spends more in his store than the national average.
Learn more about one-sample z-test here: